Problem: Evaluate $x^2y^3z$ if $x = \frac13$, $y = \frac23$, and $z = -9$.
Solution: We have  \[x^2 y^3 z = \left(\frac13\right)^2 \left(\frac23\right)^3(-9) = \frac{1}{9}\cdot \frac{8}{27}\cdot (-9) = -\frac{8}{27}\left(\frac19\cdot 9\right) = \boxed{-\frac{8}{27}}.\]